[Date Index] [Thread Index] [Author Index]

Re: Re: Radicals simplify

• To: mathgroup at smc.vnet.net
• Subject: [mg106409] Re: [mg106378] Re: [mg106361] Radicals simplify
• From: "David Park" <djmpark at comcast.net>
• Date: Mon, 11 Jan 2010 18:53:52 -0500 (EST)
• References: <201001100830.DAA05621@smc.vnet.net> <31173649.1263207631979.JavaMail.root@n11>

This can be simplified to the posters desired expression with Presentations.

Needs["Presentations`Master`"]  

(x^6*y^3)^(1/4)  
MapAt[FactorOut[x^2, Identity, CreateSubexpression], %, 1]  
Simplify[%, x >= 0] // ReleaseSubexpressions[]  

(x^6 y^3)^(1/4)
(x^4 y^3 (x^2))^(1/4)
x (x^2 y^3)^(1/4)

We factored x^2 out of the product and protected it by making it a
Subexpression. Then used Simplify and released the Subexpression.


David Park
djmpark at comcast.net
http://home.comcast.net/~djmpark/  



From: Andrzej Kozlowski [mailto:akoz at mimuw.edu.pl] 


On 10 Jan 2010, at 17:30, francix wrote:

> Hi,
> I am using Matematica 7 and need some help with Radicals.
>
> If I do
> Simplify[(x^4 y^3)^(1/4), x >= 0] I correctly have
>
> x (y^3)^(1/4)
>
> But If I do
>
> Simplify[(x^6 y^3)^(1/4), x >= 0] I get
>
> (x^6 y^3)^(1/4) and not the correct answer  x(x^2 y^3)^(1/4)
>
> Thanks in advanced.
>

The problem is that your desired formula has a higher ComplexityFunction 
than the one returned by Mathematica. Mathematica's default 
ComplexityFunction is close to LeafCount and

LeafCount[x*(x^2*y^3)^(1/4)]

13

while

LeafCount[(x^6*y^3)^(1/4)]

11

The quickest way to get the answer close to the one you want in this 
case is:

Refine[(x^6*y^3)^(1/4), x >= 0]

 x^(3/2)*(y^3)^(1/4)

To get exactly the answer you want you will need to play with 
ComplexityFunction, which in this case will be tricky.

Andrzej Kozlowski

• References:
  • Radicals simplify
    • From: "francix" <fracix@gmail.com>